LWIR FAILS to Warm the Atmosphere by Empirical Experiment.

[Crick]

They definitely hit the cold probe, that’s all!

Then what was the purpose of having an antenna? Why does anyone need an antenna? Why do astronomers build those gigantic dishes and dish arrays?

 

[jc456]

I know it was cooled,

They cooled the entire antenna? Are you sure?

I know they had to be in the middle of bumfk Egypt

New Jersey.

And yes, I have seen their radio antenna.

It's gigantic, how'd they cool it?​

Cosmic Background Radiation - Engineering and Technology History Wiki

"The CMB was discovered and measured in 1965 by radio astronomers Arno Penzias and Robert Wilson, who had been studying the “sky temperature.” Sky temperature describes the level of noise at different radio frequencies. It is required in the planning of satellite and other wireless communication systems because it enters receiving antennas along with communication signals. The measurement system devised by Penzias and Wilson consisted of antennas designed to minimize the radiation picked up from the ground, a traveling-wave maser amplifier, and helium-cooled terminations to minimize the noise level of the measurement system."

Correct:

Thy were getting interference in known reflective bands, at specific wavelengths, which are in the 2nd, 3rd, and 4th order wave length reflections. They used helium cooled instruments to remove these reflections causing noise.

You get a GOLD STAR!

You're not going to tell SSDD that the "colder" CMB photons hit the warmer antenna, are you?

He'll be heart broken.

They definitely hit the cold probe, that’s all!

They didn't hit the warm antenna first?

Is it cold?

 

[jc456]

They definitely hit the cold probe, that’s all!

Then what was the purpose of having an antenna? Why does anyone need an antenna? Why do astronomers build those gigantic dishes and dish arrays?

Looking for particular waves!

 

[Toddsterpatriot]

Cosmic Background Radiation - Engineering and Technology History Wiki

"The CMB was discovered and measured in 1965 by radio astronomers Arno Penzias and Robert Wilson, who had been studying the “sky temperature.” Sky temperature describes the level of noise at different radio frequencies. It is required in the planning of satellite and other wireless communication systems because it enters receiving antennas along with communication signals. The measurement system devised by Penzias and Wilson consisted of antennas designed to minimize the radiation picked up from the ground, a traveling-wave maser amplifier, and helium-cooled terminations to minimize the noise level of the measurement system."

Correct:

Thy were getting interference in known reflective bands, at specific wavelengths, which are in the 2nd, 3rd, and 4th order wave length reflections. They used helium cooled instruments to remove these reflections causing noise.

You get a GOLD STAR!

You're not going to tell SSDD that the "colder" CMB photons hit the warmer antenna, are you?

He'll be heart broken.

They definitely hit the cold probe, that’s all!

They didn't hit the warm antenna first?

Is it cold?

Warmer than the CMB.

 

[jc456]

Correct:

Thy were getting interference in known reflective bands, at specific wavelengths, which are in the 2nd, 3rd, and 4th order wave length reflections. They used helium cooled instruments to remove these reflections causing noise.

You get a GOLD STAR!

You're not going to tell SSDD that the "colder" CMB photons hit the warmer antenna, are you?

He'll be heart broken.

They definitely hit the cold probe, that’s all!

They didn't hit the warm antenna first?

Is it cold?

Warmer than the CMB.

Radio waves

 

[Toddsterpatriot]

You're not going to tell SSDD that the "colder" CMB photons hit the warmer antenna, are you?

He'll be heart broken.

They definitely hit the cold probe, that’s all!

They didn't hit the warm antenna first?

Is it cold?

Warmer than the CMB.

Radio waves

Radio waves

Electromagnetic Spectrum - Introduction

Yup.
Those cold photons hitting that warm antenna.

 

[Billy_Bob]

I decided to drop this post here as the information in it was used to form the basic tents of our experiment. What we observed was a firm affirmation of Dr Happer's work. I want to thank SSDD for posting it up!

Dr Happer shows why LWIR can not warm the atmosphere absent water vapor. Our experiment confirmed his findings.

"
If 99.9999999% of the IR absorbed by atmospheric CO2 is converted by molecular collisions into heat, that seems to imply that the amount of ~15 micron IR emitted by atmospheric CO2 depends only on the atmosphere's temperature (and CO2 partial pressure), not on how the air got to that temperature. [YES, I COULD HAVE SAVED A COMMENT BY READING FURTHER.] Whether the ground is very cold and emits little IR, or very warm and emits lots of IR, will not affect the amount of IR emitted by the CO2 in the adjacent atmosphere (except by affecting the temperature of that air). Is that correct? [YES, PRECISELY
"
That quote disputes your claim. Why don't you post up one that you thinks supports your claim?

Tell me Ian, if almost all absorbed LWIR is relieved by collision then there is no LWIR to be emitted fromCO2... how is your GHG atmosphere going to work now?

 

[IanC]

Tell me Ian, if almost all absorbed LWIR is relieved by collision then there is no LWIR to be emitted fromCO2... how is your GHG atmosphere going to work now?

Why are you so stupid? Why don't you understand any basic Physics? Are you asking me to explain it yet again? Two dozen times wasn't enough, why would one more iteration finally educate you?

It seems to be news to you that surface radiation energy is absorbed by CO2 and almost immediately transformed into a different form of energy that is stored in the atmosphere. I don't care that you ignored it dozens of times when I told you in the past. How has this new information changed your thinking today?

Where does this stored energy go? CO2 is already getting rid of as much energy as it can, high up where it is cold. Likewise with the other GHGs. So where does this captured surface radiation energy go? Does it just keep collecting in the atmosphere, causing higher and higher temperatures? Or does most of it return to the surface?

 

[Crick]

I'm a bit confused about something. SSDD, Billy Bob and their acolytes, for the past few weeks, have been claiming that CO2 can not warm anything because it instantaneously re-emits any photons it absorbs. Now we learn from the eminent Dr Spencer that it is a billion times more likely to lose that energy in a collision than through photon emisison. Which is it?

 

[IanC]

I'm a bit confused about something. SSDD, Billy Bob and their acolytes, for the past few weeks, have been claiming that CO2 can not warm anything because it instantaneously re-emits any photons it absorbs. Now we learn from the eminent Dr Spencer that it is a billion times more likely to lose that energy in a collision than through photon emisison. Which is it?

That's a great question!

There are a few parts to it. What is happening in a small local area (The equipartition theory). What happens to radiation energy that enters a small local area when it is more (or less) than the equilibrium produced amount. What difference (If any) does the temperature make on absorbing photons, and on emitting photons.

Cover all of those areas and you will have a strong basic understanding.

 

[Wuwei]

I'm a bit confused about something. SSDD, Billy Bob and their acolytes, for the past few weeks, have been claiming that CO2 can not warm anything because it instantaneously re-emits any photons it absorbs. Now we learn from the eminent Dr Spencer that it is a billion times more likely to lose that energy in a collision than through photon emisison. Which is it?

Relaxation time for CO2 vibration 6 microSec. In other words CO2 holds on to it's vibration state for 6 microSec, on an average.

Mean time between collisions 0.2 nsec. In other words a CO2 molecule is hit ever fraction of a nanosecond which is much sooner than a CO2 can release it's energy.

That leads to a probability of CO2 emission: = 0.2 ns / 6000 ns = 1 / 30,000

So absorption of energy of CO2 is 30,000 times higher than re-emission. This is different than what Spencer or Happer says, but the point is that it must be large.

.

 

[jc456]

I'm a bit confused about something. SSDD, Billy Bob and their acolytes, for the past few weeks, have been claiming that CO2 can not warm anything because it instantaneously re-emits any photons it absorbs. Now we learn from the eminent Dr Spencer that it is a billion times more likely to lose that energy in a collision than through photon emisison. Which is it?

That's a great question!

There are a few parts to it. What is happening in a small local area (The equipartition theory). What happens to radiation energy that enters a small local area when it is more (or less) than the equilibrium produced amount. What difference (If any) does the temperature make on absorbing photons, and on emitting photons.

Cover all of those areas and you will have a strong basic understanding.

And yet there is absolutely zero evidence that IR warms the air. You all failed

 

[Wuwei]

That's a great question!

There are a few parts to it. What is happening in a small local area (The equipartition theory). What happens to radiation energy that enters a small local area when it is more (or less) than the equilibrium produced amount. What difference (If any) does the temperature make on absorbing photons, and on emitting photons.

Cover all of those areas and you will have a strong basic understanding.

I have been working on that problem. This is what I have so far:

The following computation finds the radiating energy of CO2 in a cubic meter of air. It is in an isolated box under equilibrium. No input nor output energy is considered. The air is at ambient temperature and pressure. The basis is that the equipartition principle says 2/9 of the energy in the air will be in a vibratory mode that corresponds to 15 micron LWIR.

# of air molecules 2.53 10²⁵ / m³

# of CO2 molecules per m³ @ 400ppm = 0.0004 x 2.53 10²⁵ = 1.01 10²² /m³

The following comes from the equipartition principle.

# Degrees of freedom of CO2: 3 translation, 2 rotation, 4 vibration = 9
Two of the vibration modes are bending modes.
Probability CO2 is in a 15 micron bending mode = 2 / 9
​

# of CO2 molecules in a vibration bending mode = 2/9 x 1.01 10²² = 2.2 10²¹

Pr = Probability CO2 will radiate energy (by escaping collision)

Relaxation time for CO2 vibration 6 microSec = 6 10⁻⁶ sec
Mean time between collisions 0.2 nsec
Pr = 0.2 ns / 6000 ns = 1 / 30,000​

# of CO2 molecules radiating 15 microns every 6 microSec:

= (# vibrating) x (radiating probability) x (energy releases per sec)
= 2.2 10²¹ x 1/30,000 / (6 10⁻⁶ ) / sec = 1.22 x 10²⁵ sec⁻¹​

Photon energy at 15 microns = 1.3 10⁻²⁰ joules

Total power of 15 micron radiation per cubic meter

= (number of CO2 molecules radiating) x ( energy per CO2 photon released)
= (1.22 x 10²⁵ sec-1 ) x (1.3 10⁻²⁰ joules) = 1.59 10⁵ joules sec⁻¹ = 159 W​

That rough estimate is a large swarm of energy considering that most of the energy is transferred via the much more probable collision mode. The 159 W involves only the vibration energy of CO2 caused by the just having CO2 in the atmosphere. Any radiation input to the atmosphere would add to this.

Again, this is for a one cubic meter "box" of energy with no external energy in or out. When it comes to the open atmosphere, you have to consider an upward higher density of radiation from below entering the "box" and the "box" downward exchange radiation to the higher density atmosphere. Meanwhile the “box” will be radiating upward to a thinner density atmosphere.

That radiation exchange computation is a doable but messy integral over the spectrum of radiation paths which spans a few hundred meters. The computation is only for the radiation contribution to the atmospheric energy flow and of course ignores more important convection etc.

.

 

[SSDD]

No, you haven't, not even once. I asked for the precise equations. You're refusing to give them, as you always do. Your theory is just unsubstantiated handwaving. If you want that Nobel Prize, you're going to need to quantify things exactly. And we need to pin you down to a single solid claim, so that we can argue against it.

As if you could read an equation...laughing in your face...and all you need do is a simple search of the term IR camera with my name and you will get all the places it has been explained to you hairball...

Let's make it simpler for you, to make it harder for you to run. Forget about the specifics of the matter, as it doesn't matter what the matter is.

Its all simple hairball...which just shows how stupid you must be that you are unable to understand

We have a speck of matter at 20C.

Write the heat flow equation for that speck if the environment around it is at 19.999C, and explain how each term corresponds to which physical process.

You left out some pretty important numbers (understandable since this is all over your head) so I will have to make some assumptions about the nature of your "speck" of matter.

First, we need to know what its emissivity is...for the sake of simplicity, lets make it an ideal radiator, so its emissivity will be 1.

Second, you didn't say how large your "speck" of matter is...if you want to know how much energy it is radiating, you need to know how large it is. Again, for the sake of simplicity, lets make it a square meter...If we get much smaller than that, the radiation output of the radiator becomes vanishingly small.

0.00571 = emissivity times S-B constant times (temperature of the radiator to the 4th power minus the temperature of the surroundings to the 4th power)

If we made the radiator larger, or increased the difference between the temperature of the radiator and its surroundings, the wattage would increase.

Write the heat flow equation for that speck if the environment around it is at 20.001C, and explain how each term corresponds to which physical process.

Again..you left out some necessary information so I will have to fill in the blanks myself...First, I am going to suppose that by "surroundings" you mean something with an area such as a box...We can't apply the S-B equation to a gas as there is no area.. Again, lets make the box an ideal radiator and lets make the area of the inside of the box 6 square meters

Same as before except now the inside of the box is warmer by .001K and the inside of the box is considerably larger than the object inside... The inside of the box is radiating 0.05142W/m^2.

For the sane people, it's easy. In both cases, the equation is A * sigma * T2^4 - A * sigma * T1^4. First term is heat absorbed from environment, second term is heat radiated out.

What sort of sane S-B formula uses the SB constant twice? The proper form of the SB equation in which a radiator is radiating to cooler surroundings is: The c in Tc represents cooler as there is a basic assumption in the S-B law that the temperature of T is always greater than Tc.

Now you do it. As you say the speck behaves differently in the each environment, you'll have to have separate equations for each case.

Clearly you don't have a clue...

 

[Wuwei]

You left out some pretty important numbers (understandable since this is all over your head) so I will have to make some assumptions about the nature of your "speck" of matter.

This is your equation for a "speck" of one meter area, slightly warmer than a surround:

This is your equation for a "speck" of one meter area, slightly cooler than a surround of 6 meters.

Here is the SB equation you used:

You forgot the area, A, in your calculations.
In the first calculation the area of the "speck" was one meter, so it's OK.

In your second calculation the area should be 6 meters because A is the emitter area.

If you include the proper area you get a ludicrous result. So your whole method of using the SB equation is flawed. You get a D minus grade for your homework problem. (You get partial credit for stating the right formula. )

.

 

[Crick]

And yet there is absolutely zero evidence that IR warms the air. You all failed

You haven't been following the conversation, have you.

 

[SSDD]

You left out some pretty important numbers (understandable since this is all over your head) so I will have to make some assumptions about the nature of your "speck" of matter.

This is your equation for a "speck" of one meter area, slightly warmer than a surround:

This is your equation for a "speck" of one meter area, slightly cooler than a surround of 6 meters.

Here is the SB equation you used:

You forgot the area, A, in your calculations.
In the first calculation the area of the "speck" was one meter, so it's OK.

In your second calculation the area should be 6 meters because A is the emitter area.

If you include the proper area you get a ludicrous result. So your whole method of using the SB equation is flawed. You get a D minus grade for your homework problem. (You get partial credit for stating the right formula. )

.

Moron...the 2nd result is 6 meters...I merely forgot to change the figure for area in the equation generator. P is calculated for 6 square meters. What do you think is rediculous about that result?

As if you could possibly grade me...you fail as you didn't even recognize that P in the first equation was a smaller number than P in the second equation...nor did you recognize that P in the second equation was calculated for a radiator having an area of 6 square meters...so you caught me making an error in grammar...whoop tee do.

 

[SSDD]

And yet there is absolutely zero evidence that IR warms the air. You all failed

You haven't been following the conversation, have you.

Of course he has...there is no evidence that IR can, or does warm air.

 

[Crick]

Think about by you, Wuwei and Billy Bob over the last several days. The surface of the Earth, warmed primarily by visible light, radiates IR upward. Within a "handful of meters", it is all absorbed by CO2 and other greenhouse gases. The CO2 then loses that absorbed energy through collisions. Those collisions would be with "the air", which is then warmed.

Your assertion makes no sense. As I stated to you earlier, you have been saying in effect that "CO2 cannot warm the air because it is so rapidly warming the air."

 

[Wuwei]

Moron...the 2nd result is 6 meters...I merely forgot to change the figure for area in the equation generator. P is calculated for 6 square meters. What do you think is rediculous about that result?

That is the equation for the 1 m^2 speck emitting radiation

That is your equation for the 6 m^2 box radiating

First you have a typo and are missing a leading zero. Second, you forgot to multiply your second equation by the area of 6 m^2. If you did your answer would be

6 x 0.00571 W/m^2 = .03082 W/m^2

That is what is ridiculous. You are misusing your favorite equation.



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